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2 years ago

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A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

dimensions 5.00 cm by 5.22 cm that rotates in an adjustable magnetic field. What is the field strength needed to produce a 24.0 V peak emf

1 answer:

Sav [38]2 years ago

6 0

The field strength needed to produce a 24.0 V peak emf is 0.73T.

To find the answer, we need to know about the expression of emf.

What's the expression of peak emf produced in a rotating rectangular loops?

The peak emf produced in a rotating loops= N×B×A×w
N= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequency
So, B = emf/(N×A×w)

<h3>What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?</h3>

N= 300, A= 5cm × 5.22cm = 0.05m × 0.0522m = 0.00261 m²
Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/s
Now, B= 24/(300×0.00261×42)

B= 24/(300×0.00261×42) = 0.73T

Thus, we can conclude that the magnetic field is 0.73T.

Learn more about the electromagnetic force here:

brainly.com/question/13745767

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Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.

olganol [36]

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

$U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{x,1}$ , $K_{x,2}$ - Initial and final horizontal translational kinetic energy, measured in joules.

$K_{y,1}$ , $K_{y,2}$ - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

$m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})$

$y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}$

Where:

$y_{1}$ . $y_{2}$ - Initial and final height of the arrow, measured in meters.

$v_{y,1}$ , $v_{y,2}$ - Initial and final vertical speed of the arrow, measured in meters.

$g$ - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

$v_{y,1} = v_{1}\cdot \sin \theta$

Where:

$v_{1}$ - Magnitude of the initial velocity, measured in meters per second.

$\theta$ - Initial angle, measured in sexagesimal degrees.

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the initial vertical speed is:

$v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}$

$v_{y,1} \approx 33.352\,\frac{m}{s}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} \approx 33.352\,\frac{m}{s}$ and $v_{y,2} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{2} - y_{1} = 56.712\,m$

Second arrow

$U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}$

Where:

$U_{g,1}$ , $U_{g,3}$ - Initial and final gravitational potential energy, measured in joules.

$K_{y,1}$ , $K_{y,3}$ - Initial and final vertical translational kinetic energy, measured in joules.

$m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})$

$y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} = 82\,\frac{m}{s}$ and $v_{y,3} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{3} - y_{1} = 342.816\,m$

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

$E = U + K_{x}$

The expression is now expanded:

$E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}$

Where $v_{x}$ is the horizontal speed of the arrow, measured in meters per second.

$v_{x} = v_{1}\cdot \cos \theta$

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the horizontal speed is:

$v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}$

$v_{x} \approx 74.911\,\frac{m}{s}$

If $m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{max} = 56.712\,m$ and $v_{x} \approx 74.911\,\frac{m}{s}$ , the total mechanical energy is:

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}$

$E = 201.720\,J$

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

$E = m\cdot g \cdot y_{max}$

$m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $y_{max} = 342.816\,m$

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)$

$E = 201.720\,J$

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

7 0

3 years ago

A taut rope has a mass of 0.123 kg and a length of 3.54 m .What average power must be supplied to the rope to generate sinusoida

gladu [14]

Answer:

811.54 W

Explanation:

Solution

Begin with the equation of the time-averaged power of a sinusoidal wave on a string:

P = $\frac{1}{2\\}$ μ.T².ω².v

The amplitude is given, so we need to calculate the linear mass density of the rope, the angular frequency of the wave on the rope, and the frequency of the wave on the string.

We need to calculate the linear density to find the wave speed:

μ = $\frac{M}{L}$ = 0.123Kg/3.54m

The wave speed can be found using the linear mass density and the tension of the string:

v= 22.0 ms⁻¹

v = f/λ = 22.0/6.0×10⁻⁴

= 36666.67 s⁻¹

The angular frequency can be found from the frequency:

ω= 2πf=2π(36666.67s−1) = 2.30 ×10⁻⁵s⁻¹

Calculate the time-averaged power:

P = $\frac{1}{2}$ μΤ²×ω²×ν

= $\frac{1}{2}$ ×( 0.03475kg/m)×(0.0002)²×(2.30×10⁵)² × 22.0

= 811.54 W

5 0

3 years ago

What is one way space travel affects the human body?

Korolek [52]

Spending any length of time in microgravity can weaken bones and shrink muscles. But there are more subtle ways<span> that </span>space travel<span> wreaks havoc on the </span><span>human body</span>

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3 years ago

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Calculate the density, in g/l, of sf6 gas at 27°c and 0.500 atm pressure.

Bumek [7]

The density of the SF₆ gas at the given pressure and temperature is 2.96 g/l.

The given parameters:

<em>temperature of the SF₆ gas, T = 27 ⁰C = 273 + 27 = 300 K</em>
<em>pressure of the SF₆ gas, P = 0.5 atm</em>

The molecular mass of the SF₆ gas is calculated as follows;

M of SF₆ = 32 + (6 x 19) = 146 g/mol

The density of the SF₆ gas is calculated by applying ideal gas law as follows;

$PV = nRT\\\\PV = \frac{m}{M} RT\\\\PM = \frac{m}{V} RT\\\\PM = \rho RT\\\\\rho = \frac{PM}{RT}$

where;

<em /> $\rho$ <em> is the density of the gas</em>
<em>R is the ideal gas constant = 0.0821 L.atm/mol.K</em>

<em />

<em /> $\rho = \frac{0.5 \times 146}{0.0821 \times 300} \\\\\rho = 2.96 \ g/l$ <em />

<em />

Thus, the density of the SF₆ gas at the given pressure and temperature is 2.96 g/l.

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3 years ago

Find your average speed if you run 50m in 10s

vazorg [7]

Distance = speed / time
<span>speed = distance / time </span>

<span>so, </span>
<span>speed = 50 m / 10 sec </span>
<span>speed = 5 meters per sec</span><span />

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3 years ago

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